Question

11. The combustion analysis of 0.452 g of the hydrocarbon cubane (molecular mass, 104.16 g/mol) gives the following results: 1.528 g CO and 0.3126 g H.O. What are the empirical and molecular formulas of cubane? Empirical formula Molecular formula

Answer 1

Hydrocarbon + 02 (excesi) G (O2 + H2O Amount of CO2 produced = Amount of H2o produced = 1.528 8 0.3126 g The moleular formula of unknown hydrocarton can be calulated through following steps : Calculation of c in (0, and H in H20 : Mass of c present = 1.5289x Thea in CO2 44 g = 0.4179 Mass of H bresent - = 2 x 0.31269 X- in H2O 19 189 since one molecule H₂O contains a H atoms.so multiplied by two) 0.0347 g = moles: (ii) convert in SO Moles of Ca 0.417 9 - = 0.03475 mole Moles of H= = 0.0347 mole 0.034709 I gmolet

VDULILLA (111) Divide each molar value by lowest value. C: 0.03475 9:0347ņa H: 0.0347 0.0347 © Thus, imbipercal formula = CH Empirical mass = 13 gmole! ..in = molecular mass Empirical mass 104.16 gmolet = R 13 gmolet Thus, moleular formula = C8Hg